For your problem without any more clarification there will be an infinite number of sequences that fit your description.

You could find one by fitting a polynomial through them or use a program to generate a sequence.

Without any more constraints, a unique sequence is not really possible.

May 20th 2013, 02:34 AM

Shakarri

Re: nth term

Your numbers seem random, there has to be a pattern if you want an expression for the nth term.

May 20th 2013, 03:30 AM

ibdutt

Re: nth term

Notice that the difference between the successive terms is 7,11,15 etc and that is an AP
the sequence can be written as
6 + ( 6 + 7) + ( 6+7+11) + ( 6+7+11+15 ) and so on now i am sure you can get the nth term

May 22nd 2013, 12:00 PM

HallsofIvy

Re: nth term

I would do this using "Newton's divided difference formula". The numbers are , , , and . The "First Differences" are , , and then the "Second Differences" are , . If we assume that the sequence continues so that the second differences are alway four (and the third and higher differences are 0) we can write, in analogy with the Taylor's series, .

You will note that when n= 0, , when n= 1, , when n= 2, , and when n= 3, exactly as we want.

You could also use the fact that, since the second differences are constant, the sequence is a quadratic function of n: . Then use , and to solve for a, b, and c.